# Direction vector

## Homework Statement

2. The Celsius temperature in a region of space is given by T(x, y, z) = 2x^2 − xyz. A
particle is moving in this region with its position at time t given by x = 2t^2, y = 3t,
z = −t^2, where time and distance are measured in seconds and meters, respectively.
(a) Find the rate of change of the temperature experienced by the particle in C◦/m
when the particle is at the point P(8, 6,−4).
(b) Find the rate of change of the temperature experienced by the particle in C◦/sec
at P.

## The Attempt at a Solution

To find part a, we need to find the direction vector. However, how to find direction vector? Is velocity vector the direction vector?

If so, than part b is very easy.
Since $\frac{dT}{dt}$=$\frac{dT}{ds}\frac{ds}{dt}$
Where [/itex]=$\frac{ds}{dt}$ is the magnitute of velocity.

I wonder whether I am correct.

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
The "direction" vector is the unit tangent vector to the path. The "rate of change of T in degrees Celcius per meter" is $\nabla T\cdot \vec{v}\text{ }$ where $\vec{v}\text{ }$ is the unit tangent vector.

For part (b) at least, you do NOT need to find the "direction" or s, just use the chain rule:
$$\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial u}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$$.

That is almost the same as (a) except that you are not dividing by the length of the tangent vector to get the unit tangent vector.

The "direction" vector is the unit tangent vector to the path. The "rate of change of T in degrees Celcius per meter" is $\nabla T\cdot \vec{v}\text{ }$ where $\vec{v}\text{ }$ is the unit tangent vector.

For part (b) at least, you do NOT need to find the "direction" or s, just use the chain rule:
$$\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial u}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$$.

That is almost the same as (a) except that you are not dividing by the length of the tangent vector to get the unit tangent vector.
That means we have 2 ways to do part b?
1. Is the chain rule.
2. Just like part (a) but without dividing the length of the tangent vector?

HallsofIvy
Science Advisor
Homework Helper
Yes, the derivative of < x, y, z> with respect to t gives a "tangent vector", dividing by its length gives the unit tangent vector. The chain rule essentiall is the dot product of grad T with that non-unit tangent vector.

Probably the simplest way to do this is to do (b) first, by the chain rule, then divide that answer by the length of the vector
$$(dx/dt)\vec{i}+ (dy/dt)\vec{j}+ (dz/dt)\vec{j}$$
to get the answer to (a). That is the magnitude of the velocity vector and so is the speed with which the particle is moving. "rate of change of T as a function of time" divided by "rate of change of position as a function of time" is "rate of change of T as a function of distance. In a very loose notation:
$$\frac{\frac{dT}{dt}}{\frac{dx}{dt}}= \frac{dT}{dt}\frac{dt}{dx}= \frac{dT}{dx}$$

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